Kleinstes Palindrom nach dem Austausch
Gegeben sei eine Zeichenfolge, die einige Kleinbuchstaben und einen Sonderzeichenpunkt (.) enthält. Wir müssen alle Punkte durch ein bestimmtes Alphabetzeichen ersetzen, sodass die resultierende Zeichenfolge zu einem Palindrom wird. Bei vielen möglichen Ersetzungen müssen wir die Palindromzeichenfolge auswählen, die lexikografisch am kleinsten ist. Wenn es nach allen möglichen Ersetzungen nicht möglich ist, die Zeichenfolge in ein Palindrom umzuwandeln, wird die Ausgabe „Nicht möglich“ ausgegeben.
Beispiele:
Input : str = ab..e.c.a Output : abcaeacba The smallest palindrome which can be made after replacement is 'abcaeacba' We replaced first dot with 'c' second dot with 'a' third dot with 'a' and fourth dot with 'b' Input : str = ab..e.c.b Output : Not Possible It is not possible to convert above string into palindrome
Wir können dieses Problem wie folgt lösen. Da die resultierende Zeichenfolge ein Palindrom sein muss, können wir Paare von Nicht-Punkt-Zeichen am Anfang überprüfen. Wenn sie nicht übereinstimmen, ist eine direkte Rückkehr nicht möglich, da wir neue Zeichen nur an der Position der Punkte und nicht anderswo platzieren können.
Danach iterieren wir über die Zeichen der Zeichenfolge, wenn das aktuelle Zeichen ein Punkt ist, dann überprüfen wir das gepaarte Zeichen (Zeichen an der (n – i -1)-ten Position). Wenn dieses Zeichen ebenfalls ein Punkt ist, können wir beide Zeichen durch „a“ ersetzen, da „a“ das kleinste Kleinbuchstabenalphabet ist, das die kleinste lexikografische Zeichenfolge am Ende garantiert. Wenn Sie beide durch ein anderes Zeichen ersetzen, wird eine lexikographisch größere palindromische Zeichenfolge erstellt. In einem anderen Fall, wenn das gepaarte Zeichen kein Punkt ist, müssen wir das aktuelle Zeichen durch das gepaarte Zeichen ersetzen, um die Zeichenfolge palindromisieren zu können.
So in short If both 'i' and 'n- i- 1' are dot replace them by ‘a’ If one of them is a dot character replace that by other non-dot character
Das obige Verfahren liefert uns die lexikografisch kleinste Palindromfolge.
Durchführung:
C++ // C++ program to get lexicographically smallest // palindrome string #include using namespace std ; // Utility method to check str is possible palindrome // after ignoring . bool isPossiblePalindrome ( string str ) { int n = str . length (); for ( int i = 0 ; i < n / 2 ; i ++ ) { /* If both left and right character are not dot and they are not equal also then it is not possible to make this string a palindrome */ if ( str [ i ] != '.' && str [ n - i -1 ] != '.' && str [ i ] != str [ n - i -1 ]) return false ; } return true ; } // Returns lexicographically smallest palindrom // string if possible string smallestPalindrome ( string str ) { if ( ! isPossiblePalindrome ( str )) return 'Not Possible' ; int n = str . length (); // loop through character of string for ( int i = 0 ; i < n ; i ++ ) { if ( str [ i ] == '.' ) { // if one of character is dot replace dot // with other character if ( str [ n - i - 1 ] != '.' ) str [ i ] = str [ n - i - 1 ]; // if both character are dot then replace // them with smallest character 'a' else str [ i ] = str [ n - i - 1 ] = 'a' ; } } // return the result return str ; } // Driver code to test above methods int main () { string str = 'ab..e.c.a' ; cout < < smallestPalindrome ( str ) < < endl ; return 0 ; }
Java // Java program to get lexicographically // smallest palindrome string class GFG { // Utility method to check str is // possible palindrome after ignoring static boolean isPossiblePalindrome ( char str [] ) { int n = str . length ; for ( int i = 0 ; i < n / 2 ; i ++ ) { /* If both left and right character are not dot and they are not equal also then it is not possible to make this string a palindrome */ if ( str [ i ] != '.' && str [ n - i - 1 ] != '.' && str [ i ] != str [ n - i - 1 ] ) return false ; } return true ; } // Returns lexicographically smallest // palindrome string if possible static void smallestPalindrome ( char str [] ) { if ( ! isPossiblePalindrome ( str )) System . out . println ( 'Not Possible' ); int n = str . length ; // loop through character of string for ( int i = 0 ; i < n ; i ++ ) { if ( str [ i ] == '.' ) { // if one of character is dot // replace dot with other character if ( str [ n - i - 1 ] != '.' ) str [ i ] = str [ n - i - 1 ] ; // if both character are dot // then replace them with // smallest character 'a' else str [ i ] = str [ n - i - 1 ] = 'a' ; } } // return the result for ( int i = 0 ; i < n ; i ++ ) System . out . print ( str [ i ] + '' ); } // Driver code public static void main ( String [] args ) { String str = 'ab..e.c.a' ; char [] s = str . toCharArray (); smallestPalindrome ( s ); } } // This code is contributed // by ChitraNayal
Python 3 # Python 3 program to get lexicographically # smallest palindrome string # Utility method to check str is # possible palindrome after ignoring def isPossiblePalindrome ( str ): n = len ( str ) for i in range ( n // 2 ): # If both left and right character # are not dot and they are not # equal also then it is not possible # to make this string a palindrome if ( str [ i ] != '.' and str [ n - i - 1 ] != '.' and str [ i ] != str [ n - i - 1 ]): return False return True # Returns lexicographically smallest # palindrome string if possible def smallestPalindrome ( str ): if ( not isPossiblePalindrome ( str )): return 'Not Possible' n = len ( str ) str = list ( str ) # loop through character of string for i in range ( n ): if ( str [ i ] == '.' ): # if one of character is dot # replace dot with other character if ( str [ n - i - 1 ] != '.' ): str [ i ] = str [ n - i - 1 ] # if both character are dot # then replace them with # smallest character 'a' else : str [ i ] = str [ n - i - 1 ] = 'a' # return the result return str # Driver code if __name__ == '__main__' : str = 'ab..e.c.a' print ( '' . join ( smallestPalindrome ( str ))) # This code is contributed by ChitraNayal
C# // C# program to get lexicographically // smallest palindrome string using System ; public class GFG { // Utility method to check str is // possible palindrome after ignoring static bool isPossiblePalindrome ( char [] str ) { int n = str . Length ; for ( int i = 0 ; i < n / 2 ; i ++ ) { /* If both left and right character are not dot and they are not equal also then it is not possible to make this string a palindrome */ if ( str [ i ] != '.' && str [ n - i - 1 ] != '.' && str [ i ] != str [ n - i - 1 ]) return false ; } return true ; } // Returns lexicographically smallest // palindrome string if possible static void smallestPalindrome ( char [] str ) { if ( ! isPossiblePalindrome ( str )) Console . WriteLine ( 'Not Possible' ); int n = str . Length ; // loop through character of string for ( int i = 0 ; i < n ; i ++ ) { if ( str [ i ] == '.' ) { // if one of character is dot // replace dot with other character if ( str [ n - i - 1 ] != '.' ) str [ i ] = str [ n - i - 1 ]; // if both character are dot // then replace them with // smallest character 'a' else str [ i ] = str [ n - i - 1 ] = 'a' ; } } // return the result for ( int i = 0 ; i < n ; i ++ ) Console . Write ( str [ i ] + '' ); } // Driver code public static void Main () { String str = 'ab..e.c.a' ; char [] s = str . ToCharArray (); smallestPalindrome ( s ); } } // This code is contributed by PrinciRaj1992
PHP // PHP program to get lexicographically // smallest palindrome string // Utility method to check str is // possible palindrome after ignoring function isPossiblePalindrome ( $str ) { $n = strlen ( $str ); for ( $i = 0 ; $i < $n / 2 ; $i ++ ) { /* If both left and right character are not dot and they are not equal also then it is not possible to make this string a palindrome */ if ( $str [ $i ] != '.' && $str [ $n - $i - 1 ] != '.' && $str [ $i ] != $str [ $n - $i - 1 ]) return false ; } return true ; } // Returns lexicographically smallest // palindrome string if possible function smallestPalindrome ( $str ) { if ( ! isPossiblePalindrome ( $str )) return 'Not Possible' ; $n = strlen ( $str ); // loop through character of string for ( $i = 0 ; $i < $n ; $i ++ ) { if ( $str [ $i ] == '.' ) { // if one of character is dot // replace dot with other character if ( $str [ $n - $i - 1 ] != '.' ) $str [ $i ] = $str [ $n - $i - 1 ]; // if both character are dot // then replace them with // smallest character 'a' else $str [ $i ] = $str [ $n - $i - 1 ] = 'a' ; } } // return the result return $str ; } // Driver code $str = 'ab..e.c.a' ; echo smallestPalindrome ( $str ); // This code is contributed // by ChitraNayal ?>
JavaScript < script > // Javascript program to get lexicographically // smallest palindrome string // Utility method to check str is // possible palindrome after ignoring function isPossiblePalindrome ( str ) { let n = str . length ; for ( let i = 0 ; i < Math . floor ( n / 2 ); i ++ ) { /* If both left and right character are not dot and they are not equal also then it is not possible to make this string a palindrome */ if ( str [ i ] != '.' && str [ n - i - 1 ] != '.' && str [ i ] != str [ n - i - 1 ]) return false ; } return true ; } // Returns lexicographically smallest // palindrome string if possible function smallestPalindrome ( str ) { if ( ! isPossiblePalindrome ( str )) document . write ( 'Not Possible' ); let n = str . length ; // loop through character of string for ( let i = 0 ; i < n ; i ++ ) { if ( str [ i ] == '.' ) { // if one of character is dot // replace dot with other character if ( str [ n - i - 1 ] != '.' ) str [ i ] = str [ n - i - 1 ]; // if both character are dot // then replace them with // smallest character 'a' else str [ i ] = str [ n - i - 1 ] = 'a' ; } } // return the result for ( let i = 0 ; i < n ; i ++ ) document . write ( str [ i ] + '' ); } // Driver code let str = 'ab..e.c.a' ; let s = str . split ( '' ); smallestPalindrome ( s ); // This code is contributed by rag2127 < /script>
Ausgabe
abcaeacba
Zeitkomplexität: O(n) wobei n die Länge der Zeichenfolge ist.
Komplexität des Hilfsraums: O(1)
